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lgbasallote
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Prove that if n is a perfect square, then n +2 is not a perfect square
 2 years ago
 2 years ago
lgbasallote Group Title
Prove that if n is a perfect square, then n +2 is not a perfect square
 2 years ago
 2 years ago

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lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i suppose first step would be to assume n is perfect square
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
so n = m^2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
then i suppose i find n^2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i mean n + 2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
n + 2 would be m^2 + 2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
so now... i assume m^2 + 2 is a perfect square...
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
we have no 2 perfect square numbers with difference 2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
so i let m^2 + 2 = k^2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
@mukushla that's not really a proof.....
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
you may continue @igba
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
hmm i suppose the next step would be m^2  k^2 = 2 so (m+k)(mk) = 2 i don't think this proves anything...
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
actually that is :) \[m^2k^2=2\]\[(mk)(m+k)=2\]and this is impossible
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
why so?
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
well,,you've almost proved it.. now m and k both are integers/.
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
and its k^2 m^2 = 2 and not otherwise..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
2 = (k + m)(km)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
now...integers huh...
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
\[m+k>mk\]so\[m+k=2\]so\[m=k=1\]but it gives\[mk=0\]
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
you know k>m.. so from (k+m)(km)= 1*2 =>k+m =2 and km = 1 this doesnt have an integral solution..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
why do i know k > m again?
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
m^2 + 2 = k^2 both m and k are positive integers hence k>m
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
in layman language,, k^2  m^2 >0
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
when was it assumed that m and k are positive?
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
what else do you mean by perfect square ?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i don't see how that relates.... the definition of a perfect square is x = k^2
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
4 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
@mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
(2)^2 is also 4...
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
you take the absolute value
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
i mean you have to..
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
see,, (k+m)(km) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take ve values,, negative * negative = positive ,,i.e. treated as absolute value..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
...i don't think this is a valid proof....
 2 years ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
another explanation might me x is a perfect square because its y^2 x = y^2 = (y)^2 sqrt(x) =  y  =  y 
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
since when did sqrt x become y
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
so and i think that this difference of two will be the ,,key" of this proof of ...
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
@lgbasallote what is your opinion from this ,please ?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
my opinion is that this is some tricky algebra
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
how do you think it ???
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
so and why ?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
because i can't think of a proof
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
how can you prove/disprove that the difference of two squares is 2...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
hmmm...wait...
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
their fields of specialty aren't exactly in proving
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
What's wrong with my proof?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
anyway... m^2 + 2 = k^2 k^2  m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2  (2y^2) = 2 4x^2  4y^2 = 2 then... 2x^2  2y^2 = 1 x^2  y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
@estudier i mean
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
@mukushla too
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
So, that doesn't mean it's invalid..
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
it does actually.....you can't do infinite substitutions to prove anything....
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
There is no infinity involved, all integer squares end in the numbers I gave.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
infinite loops involves infinity agree?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
No infinite loop either, just logic.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
An integer ends in one of digits 0 to 9. The squares must end in etc....
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
but we both know that setting values is illegal in proving
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
I am not setting any values.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
you're checking the squares of each number
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
and though there is a pattern...there is still susbtitution involved
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
and then you'll have to substitute forever to verify that the pattern really does not break
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
then infinity is involved
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
so you do admit that you checked the 0 to 9 squares
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Anything else is impossible.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
...that's substitution
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
anyway...my point is...substitution isn't really allowed in proving
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
You will find that this fact is used a lot in many valid proofs in number theory.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
perhaps
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
can you prove it? that it's a fact about squares?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Yes, all numbers end in 0 to 9 and so their squares end in .....QED.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
you didn't really demonstrate anything...you just repeated what you said
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
You asked for proof, I just gave u a proof.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
If you deny my proof, provide a counterexample.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
since you try so hard to defend your proof, i'll accept it
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
That's very gracious, thank you:)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
do check on my algebra proof though please. i would like to know if that's the way contradiction works
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
OK, I will look at it now....
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
again i ask... how is k > m
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
(k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[km=1\]\[k+m=2\]add them\[2k=3\] Contradiction
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
again i ask for the nth time @mukushla how is k > m
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
ok man i just want to answer ur question
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
"2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
well because 8  4 is not 2 but it's even minus even
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
@mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
OK, so finally, the algebra Given n=k^2 > (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
i thought they are positive :) sorry
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
why (k+1)^2 @estudier ?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
I already, said, that's the next square....
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
next square?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
is this still related to my proof?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
? n is k^2, what is the next square?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
no problem ... this is better\[k^2m^2=2\]\[(km)(k+m)=2\]so\[km=1\]\[k+m=2\]because this time im sure that \(k>m\) ...now add them\[2k=3\]Contradiction and we are done
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
No, this proof Given n=k^2 > (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
are you allowed to do that? @mukushla
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
why not...for any real number \[a^2=a^2\]
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Given n=k^2 > (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
That's 2 proofs I have given now...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Think about it....:)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
in the meanwhile... @mukushla why k  m = 1?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
because\[km<k+m\]
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
how does that make k  m = 1
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
u have 2 positive integers with ab=2 and a<b what are a and b....clearly 1 and 2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
who said those are positive?
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
k>m right?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
something feels wrong...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
anyway...where did you get 2k > 3
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
ok...think on what i said...because i believe thats the shortest way to prove ur statement
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
2k = 3 *
 2 years ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
adding\[km=1\]and\[k+m=2\]
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Going back to your question about 2, u can just adjust it to difference of 2q...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
im really skeptical about this... especially that k^2  m^2 = 2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
difference of 2q?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
You were worried about the validity of your statement, change it to 2q instead of just 2.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Or 2p, or whatever letter u like best.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
1) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
oh. when i said difference of two numbers equal to 2
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
and @mukushla how did you get k > m
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Yes, the algebra is completely general...
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
...which is greater than n+3, right?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Number theorists don't pay attention to 0
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
unless k is 0....
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
why not?
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Can I try?
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
More the merrier.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
"because i know nothing about it" But you do now:)
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
it's not like it's limited to the 99s....
 2 years ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.
 2 years ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
Yes, that's similar to the one I gave...
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
"number theorists don't consider 0" http://en.wikipedia.org/wiki/Natural_number
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.3
http://mathforum.org/library/drmath/view/63510.html Using 0,1,4, etc...
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?
 2 years ago

jhonyy9 Group TitleBest ResponseYou've already chosen the best response.0
so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 2 so m1,2 = +/ sqrt(x^2 2)
 2 years ago
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